Chapter 21Electric Charge

1 22:16

1. Electric Charge

Electric charge is an intrinsic characteristic of the fundamental particles making upobjects. It is a property that comes automatically with those particles wheneverthey exist.

The vast amount of charge in an everyday object is hidden because the objectcontains equal amounts of positive charge and negative charge. Such an object issaid to be electrically neutral; it contains no net charge.

2 22:16

1. Electric Charge

Charges with the same electrical sign repel each other, and charges with oppositeelectrical signs attract each other.3 22:16

2. Conductors and Insulators

Conductors are materials through which charge can move rather feely. Examples ofconductors include metals, tap water and human body.

Nonconductors or insulators are martials though which charge cannot move freely.Examples of insulators include rubber, glass and distilled water.

Semiconductors are materials that are intermediate between conductors andinsulators. Examples of semiconductors include silicon and germanium.

Superconductors are perfect conductors.

4 22:16

2. Conductors and Insulators

C and D attract.

B and D attract.

5 22:16

−

− −

−

−+

+ + +

−

− −

3. Coulomb’s Law

The repulsion or attraction force between twocharges, due to their charges, is called anelectrostatic force. The electrostatic force is givenby Coulombs' law.

The force on particle 1 in the figure is given by

𝐹 = 𝑘𝑞1𝑞2

𝑟2 𝑟.

where 𝑟 is a unit vector along the axis extendingthrough the two particles, 𝑟 is the distancebetween them, and 𝑘 is the electrostatic constant.

6 22:16

3. Coulomb’s Law

The SI unit of charge is the coulomb (C). The coulomb unit is derived from the SIbase unit ampere for the electric current 𝑖. Current is the rate at which chargemoves past a point or through a region:

𝑖 =𝑑𝑞

𝑑𝑡.

This relation tells us that

1 C = 1 A 1 s .

The electrostatic constant 𝑘 is usually written 1/4𝜋𝜀0. The magnitude of the forcein Coulomb’s law becomes

𝐹 =1

4𝜋𝜀0

𝑞1𝑞2

𝑟2.

7 22:16

3. Coulomb’s Law

The electrostatic constant 𝑘 has the value

𝑘 =1

4𝜋𝜀0= 8.99 × 109

N ∙ m2

C2.

The constant 𝜀0 is called the permittivity constant and has the value

𝜀0 = 8.85 × 10−12C2

N ∙ m2.

8 22:16

3. Coulomb’s Law

If we have 𝑛 charged particles, they interact independently in pairs and the forceon any one of them (say particle 1), is given by the vector sum

𝐹1,net = 𝐹12 + 𝐹13 + 𝐹14 + ⋯ + 𝐹1𝑛.

The shell theorem for electrostatics:

• A shell of uniform charge attracts or repels a charged particle that is outside theshell as if all the shell’s charge were concentrated at its center.

• If a charged particle is located inside a shell of uniform charge, there is no netelectrostatic force on the particle from the shell.

9 22:16

3. Coulomb’s Law

Spherical Conductors:

When excess charge exists on a spherical shell, it spreads uniformly over theexternal surface of the sphere. This is due to the fact that charges of the same signrepel each other and try to redistribute themselves to maximize the distancesbetween them.

10 22:16

3. Coulomb’s Law

(a) Leftward.

(b) Leftward.

(c) Leftward.

11 22:16

3. Coulomb’s Law

Example 1:

(a) The figure shows two positively chargedparticles fixed in place on an 𝑥 axis. The chargesare 𝑞1 = 1.60 × 10−19 C and 𝑞2 = 3.20× 10−19 C, and the particle separation is 𝑅0

= 0.0200 m . What are the magnitude anddirection of the electrostatic force on particle 1from particle 2?

𝐹12 =1

4𝜋𝜀0

𝑞1 𝑞2

𝑅2.

12 22:16

3. Coulomb’s Law

𝐹12

=1.60 × 10−19 C 3.20 × 10−19 C

4𝜋 8.85 × 10−12 ∙C2

N ∙ m2 0.0200 m 2

= 1.15 × 10−24 N.

𝐹12 is along the negative 𝑥 axis. In unit-vectornotation

𝐹12 = − 1.15 × 10−24 N i.

13 22:16

3. Coulomb’s Law

(b) The figure is identical to previous exceptthat particle 3 now lies on the 𝑥 axis betweenparticles 1 and 2. Particle 3 has charge 𝑞3= −3.20 × 10−19 C and is at a distance 3/4𝑅from particle 1. What is the net electrostaticforce 𝐹1,net on particle 1 due to particles 2 and3?

𝐹1,net = 𝐹1,2 + 𝐹1,3

𝐹13 =1

4𝜋𝜀0

𝑞1 𝑞3

3𝑅/4 2

14 22:16

3. Coulomb’s Law

𝐹13 =1

4𝜋𝜀0

𝑞1 𝑞3

3𝑅/4 2

𝐹13 =1.60 × 10−19 C 3.20 × 10−19 C

4𝜋 8.85 × 10−12 ∙C2

N ∙ m234

2

0.0200 m 2

= 2.05 × 10−24 N.

𝐹13 = 2.05 × 10−24 N i.

𝐹1,net = 𝐹12 + 𝐹13

= − 1.15 × 10−24 N i + 2.05 × 10−24 N i

= 9.0 × 10−25 N i.15 22:16

3. Coulomb’s Law

(c) The figure is identical to that in part (a)except that particle 4 is now included. It hascharge 𝑞4 = −3.20 × 10−19 C, is at a distance 3/4𝑅 from particle 1, and lies on a line thatmakes an angle 𝜃 = 60° with the 𝑥 axis. What isthe net electrostatic force on particle 1 due toparticles 2 and 4?

𝐹1,net = 𝐹1,2 + 𝐹1,4

𝐹14 =1

4𝜋𝜀0

𝑞1 𝑞4

3𝑅/4 2= 2.05 × 10−24 N.

𝐹1,4 = 𝐹14 cos 60° i + 𝐹14 sin 60° J

16 22:16

3. Coulomb’s Law

𝐹1,4 = 1.025 × 10−24 N i + 1.775 × 10−24 N J.

𝐹1,net = 𝐹1,2 + 𝐹1,4 = − 1.15 × 10−24 N i

+ 1.025 × 10−24 N i + 1.775 × 10−24 N J

= − 1.25 × 10−25 N i + 1.775 × 10−24 N J.

𝐹1,net = 𝐹1,net,𝑥2 + 𝐹1,net,𝑦

2

= 1.025 × 10−24 N.

𝜙 = tan−1𝐹1,net,𝑦

𝐹1,net,𝑥= −86° + 180° = 94°.

17 22:16

3. Coulomb’s Law

(A) a, c, then b.

(B) Less.

18 22:16

3. Coulomb’s Law

Example 2: The figure shows two particles fixedin place: a particle of charge 𝑞1 = +8𝑞 at theorigin and a particle of charge 𝑞2 = −2𝑞 at 𝑥= 𝐿. At what point (other than infinitely faraway) can a proton be placed so that it is inequilibrium (the net force on it is zero)?

19 22:16

3. Coulomb’s Law

Example 2: The figure shows two particles fixedin place: a particle of charge 𝑞1 = +8𝑞 at theorigin and a particle of charge 𝑞2 = −2𝑞 at 𝑥= 𝐿. At what point (other than infinitely faraway) can a proton be placed so that it is inequilibrium (the net force on it is zero)?

The proton must be placed to right of 𝑞2.

We want to have that

𝐹1 = 𝐹2.

Coulomb’s law gives

20 22:16

3. Coulomb’s Law

Coulomb’s law gives

1

4𝜋𝜀0

𝑞1𝑞𝑝

𝑥2=

1

4𝜋𝜀0

𝑞2𝑞𝑝

(𝑥 − 𝐿)2,

or

1

4𝜋𝜀0

8𝑞𝑞𝑝

𝑥2=

1

4𝜋𝜀0

2𝑞𝑞𝑝

(𝑥 − 𝐿)2.

Simplifying gives𝑥2

(𝑥 − 𝐿)2= 4,

which gives 𝑥 = 2𝐿.

21 22:16

3. Coulomb’s Law

Example 3: In figure (a) two identical,electrically isolated conducting spheres 𝐴 and 𝐵are separated by a (center-to-center) distance 𝑎that is large compared to the spheres. Sphere 𝐴has a positive charge of 𝑄, and sphere 𝐵 iselectrically neutral. Initially, there is noelectrostatic force between the spheres.

(a) Suppose the spheres are connected for amoment by a conducting wire. The wire is thinenough so that any net charge on it isnegligible. What is the electrostatic forcebetween the spheres after the wire is removed?

22 22:16

3. Coulomb’s Law

Because the spheres are identical, connecting them means that they end up with identical charges. Coulomb’s law gives

𝐹 =1

4𝜋𝜀0

𝑞𝐴 𝑞𝐵

𝑅2=

1

4𝜋𝜀0

𝑄/2 𝑄/2

𝑎2

=𝑄2

16𝜋𝜀0𝑎2.

23 22:16

3. Coulomb’s Law

(b) Next, suppose sphere 𝐴 is groundedmomentarily, and then the ground connectionis removed. What now is the electrostatic forcebetween the spheres?

24 22:16

3. Coulomb’s Law

(b) Next, suppose sphere 𝐴 is groundedmomentarily, and then the ground connectionis removed. What now is the electrostatic forcebetween the spheres?

The electrostatic force between the spheres isnow zero.

25 22:16

4. Charge is Quantized and Conserved

Any charge 𝑞 (positive or negative) can be written as

𝑞 = 𝑛𝑒, 𝑛 = ±1, ±2, ±3, … ,

where 𝑒 is the elementary charge, which has the approximate value

𝑒 = 1.602 × 10−19 C.

We say that a physical quantity is quantized if it can have only discrete values. Anobject can have charge of +10𝑒, −32𝑒, but not 4.32𝑒.

Charge Conservation

The net charge of any isolated system cannot change.

26 22:16

4. Charge is Quantized and Conserved

After the spheres touch, their net charge is −30𝑒 and the charge of each sphere is− 15𝑒.

27 22:16

4. Charge is Quantized and Conserved

Example 4: The nucleus in an iron atom has a radius of about 4.0 × 10−15 m andcontains 26 protons.

(a) What is the magnitude of the repulsive electrostatic force between two of theprotons that are separated by 4.0 × 10−15 m?

𝐹 = 𝑘𝑞𝐴 𝑞𝐵

𝑅2= 8.99 × 109 N ∙ m2/C2

1.602 × 10−19 C 2

4.0 × 10−15 m 2= 14 N.

(b) What is the magnitude of the gravitational force between those same twoprotons?

𝐹𝑔 = 𝐺𝑚𝑝

2

𝑅2= 6.67 × 10−11 N ∙ m2/kg2

1.67 × 10−27 kg 2

4.0 × 10−15 m 2= 1.2 × 10−35 N.

28 22:16